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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

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  • Volume of the balloon $V=\large\frac{4}{3}$$\pi r^3$
Step 1:
Volume of the balloon $V=\large\frac{4}{3}$$\pi r^3$
Diff w.r.t $t$ we get,
$\large\frac{dv}{dt}=\frac{4}{3}$$\pi 3r^2\large\frac{dr}{dt}$
When $t=0,r=$3 units
When $t=3,r=$6 units
It is given $\large\frac{dv}{dt}$=constant=k(say)
$\therefore k=4\pi r^2\large\frac{dr}{dt}$
Separating the variables we get,
$\therefore kdt=4\pi r^2dr$
Step 2:
Integrate on both sides
$k\int dt=4\pi\int r^2dr$
$kt=4\pi \large\frac{r^3}{3}$$+c$------(1)
When $t=0,r=$3units
$\Rightarrow k(0)=\large\frac{4\pi(3)^3}{3}$$+c$
$c=36\pi$
Step 3:
When $t=3sec,r=6$units
$\Rightarrow k(3)=\large\frac{4\pi(6)^3}{3}$$+36\pi$
$3k=288\pi+36\pi$
$k=\large\frac{324\pi}{3}$
$\;\;\;=108\pi$
$\therefore$ Hence $c=36\pi$ and $k=108\pi$
Step 4:
Now substituting the values in equ(1)
$108\pi t=\large\frac{4}{3}$$\pi r^3+36\pi$
$4\pi r^3=36\pi-108\pi t$
$r^3=9-27t$
$\therefore r=\sqrt[\large 3]{9-27t}$
answered Nov 8, 2013 by sreemathi.v
edited Nov 8, 2013 by sreemathi.v
 

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